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construct circle defined by point, radius and enclosed & touched circle
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If this can easily be derived from an already answered question please be so kind to point me there. I tried but couldn't find an equivalent problem.
I have one circle $D$ with the origin as center fully defined by its radius $r_D$.
I have a given point $P$ which is outside $D$.
Now I want to construct a circle $H$ with known radius $r_H > r_D$, such that $P$ is on the circle and $D$ is enclosed inside $H$ and $H$ is touching $D$.
Where is the circle center of $H$/how do I calculate the point where $D=H$?
There are two solutions (though I just need one).
geometry circle
edited Jul 30 at 17:30
KReiser
7,45011230
asked Jul 30 at 13:10
the_toast
82
add a comment |Â
up vote
1
down vote
favorite
If this can easily be derived from an already answered question please be so kind to point me there. I tried but couldn't find an equivalent problem.
I have one circle $D$ with the origin as center fully defined by its radius $r_D$.
I have a given point $P$ which is outside $D$.
Now I want to construct a circle $H$ with known radius $r_H > r_D$, such that $P$ is on the circle and $D$ is enclosed inside $H$ and $H$ is touching $D$.
Where is the circle center of $H$/how do I calculate the point where $D=H$?
There are two solutions (though I just need one).
geometry circle
edited Jul 30 at 17:30
KReiser
7,45011230
asked Jul 30 at 13:10
the_toast
82
connect center of $D$ (let's say point $O$) with $P$. Now draw a perpendicular to $OP$ that goes thru $O$. The perpendicular $OA$ will intersect $D$ at two points $A$ and $B$ (hence two solutions). Triange $OPA$ is a right triangle. Find $angleOPA$. Construct angle equal to $90-2 cdot OPA$ that has $P$ as vertex and $OP$ as one of the sides (again, there are two ways to construct the angle, choose wisely), The other side of the angle will intersect line $OA$ at some point, that point will be the center of $H$.
– Vasya
Jul 30 at 13:35
Hi, Thanks for your suggestion. The circle center of $H$ cannot be on the line $OA$ though. Your suggestion sets $r_H=3*r_D$ (if I'm not mistaken) but the radius $r_H$ is actually known and fixed.
– the_toast
Jul 30 at 14:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If this can easily be derived from an already answered question please be so kind to point me there. I tried but couldn't find an equivalent problem.
I have one circle $D$ with the origin as center fully defined by its radius $r_D$.
I have a given point $P$ which is outside $D$.
Now I want to construct a circle $H$ with known radius $r_H > r_D$, such that $P$ is on the circle and $D$ is enclosed inside $H$ and $H$ is touching $D$.
Where is the circle center of $H$/how do I calculate the point where $D=H$?
There are two solutions (though I just need one).
geometry circle
edited Jul 30 at 17:30
KReiser
7,45011230
asked Jul 30 at 13:10
the_toast
82
If this can easily be derived from an already answered question please be so kind to point me there. I tried but couldn't find an equivalent problem.
I have one circle $D$ with the origin as center fully defined by its radius $r_D$.
I have a given point $P$ which is outside $D$.
Now I want to construct a circle $H$ with known radius $r_H > r_D$, such that $P$ is on the circle and $D$ is enclosed inside $H$ and $H$ is touching $D$.
Where is the circle center of $H$/how do I calculate the point where $D=H$?
There are two solutions (though I just need one).
geometry circle
edited Jul 30 at 17:30
KReiser
7,45011230
asked Jul 30 at 13:10
the_toast
82
edited Jul 30 at 17:30
KReiser
7,45011230
edited Jul 30 at 17:30
KReiser
7,45011230
edited Jul 30 at 17:30
KReiser
7,45011230
7,45011230
asked Jul 30 at 13:10
the_toast
82
asked Jul 30 at 13:10
the_toast
82
asked Jul 30 at 13:10
the_toast
82
82
connect center of $D$ (let's say point $O$) with $P$. Now draw a perpendicular to $OP$ that goes thru $O$. The perpendicular $OA$ will intersect $D$ at two points $A$ and $B$ (hence two solutions). Triange $OPA$ is a right triangle. Find $angleOPA$. Construct angle equal to $90-2 cdot OPA$ that has $P$ as vertex and $OP$ as one of the sides (again, there are two ways to construct the angle, choose wisely), The other side of the angle will intersect line $OA$ at some point, that point will be the center of $H$.
– Vasya
Jul 30 at 13:35
Hi, Thanks for your suggestion. The circle center of $H$ cannot be on the line $OA$ though. Your suggestion sets $r_H=3*r_D$ (if I'm not mistaken) but the radius $r_H$ is actually known and fixed.
– the_toast
Jul 30 at 14:11
add a comment |Â
connect center of $D$ (let's say point $O$) with $P$. Now draw a perpendicular to $OP$ that goes thru $O$. The perpendicular $OA$ will intersect $D$ at two points $A$ and $B$ (hence two solutions). Triange $OPA$ is a right triangle. Find $angleOPA$. Construct angle equal to $90-2 cdot OPA$ that has $P$ as vertex and $OP$ as one of the sides (again, there are two ways to construct the angle, choose wisely), The other side of the angle will intersect line $OA$ at some point, that point will be the center of $H$.
– Vasya
Jul 30 at 13:35
Hi, Thanks for your suggestion. The circle center of $H$ cannot be on the line $OA$ though. Your suggestion sets $r_H=3*r_D$ (if I'm not mistaken) but the radius $r_H$ is actually known and fixed.
– the_toast
Jul 30 at 14:11
connect center of $D$ (let's say point $O$) with $P$. Now draw a perpendicular to $OP$ that goes thru $O$. The perpendicular $OA$ will intersect $D$ at two points $A$ and $B$ (hence two solutions). Triange $OPA$ is a right triangle. Find $angleOPA$. Construct angle equal to $90-2 cdot OPA$ that has $P$ as vertex and $OP$ as one of the sides (again, there are two ways to construct the angle, choose wisely), The other side of the angle will intersect line $OA$ at some point, that point will be the center of $H$.
– Vasya
Jul 30 at 13:35
connect center of $D$ (let's say point $O$) with $P$. Now draw a perpendicular to $OP$ that goes thru $O$. The perpendicular $OA$ will intersect $D$ at two points $A$ and $B$ (hence two solutions). Triange $OPA$ is a right triangle. Find $angleOPA$. Construct angle equal to $90-2 cdot OPA$ that has $P$ as vertex and $OP$ as one of the sides (again, there are two ways to construct the angle, choose wisely), The other side of the angle will intersect line $OA$ at some point, that point will be the center of $H$.
– Vasya
Jul 30 at 13:35
Hi, Thanks for your suggestion. The circle center of $H$ cannot be on the line $OA$ though. Your suggestion sets $r_H=3*r_D$ (if I'm not mistaken) but the radius $r_H$ is actually known and fixed.
– the_toast
Jul 30 at 14:11
Hi, Thanks for your suggestion. The circle center of $H$ cannot be on the line $OA$ though. Your suggestion sets $r_H=3*r_D$ (if I'm not mistaken) but the radius $r_H$ is actually known and fixed.
– the_toast
Jul 30 at 14:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Notice that $H$ must touch (i.e. not intersect) $D$. This means it's center is located on a circle around the center of $D = (0,0)$ with the radius $r_H - r_D$. Simultaniously it must be on a circle with radius $r_D$ with center $P$.
Now your problem is reduced to finding the intersection points of two circles.
Say: $(c_x,c_y)$ is the center of $H$ and wlog $P = (P_x,0)$.
You have the equations:
$$
c_x^2+c_y^2 = (r_D - r_H)^2 text and (c_x - P_x)^2+c_y^2 = r_H^2
$$
The rest should be simple, i guess.
answered Jul 30 at 14:12
denklo
913
This is a beautifully simple solution that none of the 5 people I asked came up with. Props!
– the_toast
Jul 30 at 14:17
Glad i could help!
– denklo
Jul 30 at 14:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Notice that $H$ must touch (i.e. not intersect) $D$. This means it's center is located on a circle around the center of $D = (0,0)$ with the radius $r_H - r_D$. Simultaniously it must be on a circle with radius $r_D$ with center $P$.
Now your problem is reduced to finding the intersection points of two circles.
Say: $(c_x,c_y)$ is the center of $H$ and wlog $P = (P_x,0)$.
You have the equations:
$$
c_x^2+c_y^2 = (r_D - r_H)^2 text and (c_x - P_x)^2+c_y^2 = r_H^2
$$
The rest should be simple, i guess.
answered Jul 30 at 14:12
denklo
913
This is a beautifully simple solution that none of the 5 people I asked came up with. Props!
– the_toast
Jul 30 at 14:17
Glad i could help!
– denklo
Jul 30 at 14:28
add a comment |Â
up vote
1
down vote
accepted
Notice that $H$ must touch (i.e. not intersect) $D$. This means it's center is located on a circle around the center of $D = (0,0)$ with the radius $r_H - r_D$. Simultaniously it must be on a circle with radius $r_D$ with center $P$.
Now your problem is reduced to finding the intersection points of two circles.
Say: $(c_x,c_y)$ is the center of $H$ and wlog $P = (P_x,0)$.
You have the equations:
$$
c_x^2+c_y^2 = (r_D - r_H)^2 text and (c_x - P_x)^2+c_y^2 = r_H^2
$$
The rest should be simple, i guess.
answered Jul 30 at 14:12
denklo
913
This is a beautifully simple solution that none of the 5 people I asked came up with. Props!
– the_toast
Jul 30 at 14:17
Glad i could help!
– denklo
Jul 30 at 14:28
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Notice that $H$ must touch (i.e. not intersect) $D$. This means it's center is located on a circle around the center of $D = (0,0)$ with the radius $r_H - r_D$. Simultaniously it must be on a circle with radius $r_D$ with center $P$.
Now your problem is reduced to finding the intersection points of two circles.
Say: $(c_x,c_y)$ is the center of $H$ and wlog $P = (P_x,0)$.
You have the equations:
$$
c_x^2+c_y^2 = (r_D - r_H)^2 text and (c_x - P_x)^2+c_y^2 = r_H^2
$$
The rest should be simple, i guess.
answered Jul 30 at 14:12
denklo
913
Notice that $H$ must touch (i.e. not intersect) $D$. This means it's center is located on a circle around the center of $D = (0,0)$ with the radius $r_H - r_D$. Simultaniously it must be on a circle with radius $r_D$ with center $P$.
Now your problem is reduced to finding the intersection points of two circles.
Say: $(c_x,c_y)$ is the center of $H$ and wlog $P = (P_x,0)$.
You have the equations:
$$
c_x^2+c_y^2 = (r_D - r_H)^2 text and (c_x - P_x)^2+c_y^2 = r_H^2
$$
The rest should be simple, i guess.
answered Jul 30 at 14:12
denklo
913
answered Jul 30 at 14:12
denklo
913
answered Jul 30 at 14:12
denklo
913
answered Jul 30 at 14:12
denklo
913
913
This is a beautifully simple solution that none of the 5 people I asked came up with. Props!
– the_toast
Jul 30 at 14:17
Glad i could help!
– denklo
Jul 30 at 14:28
add a comment |Â
This is a beautifully simple solution that none of the 5 people I asked came up with. Props!
– the_toast
Jul 30 at 14:17
Glad i could help!
– denklo
Jul 30 at 14:28
This is a beautifully simple solution that none of the 5 people I asked came up with. Props!
– the_toast
Jul 30 at 14:17
This is a beautifully simple solution that none of the 5 people I asked came up with. Props!
– the_toast
Jul 30 at 14:17
Glad i could help!
– denklo
Jul 30 at 14:28
Glad i could help!
– denklo
Jul 30 at 14:28
add a comment |Â
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connect center of $D$ (let's say point $O$) with $P$. Now draw a perpendicular to $OP$ that goes thru $O$. The perpendicular $OA$ will intersect $D$ at two points $A$ and $B$ (hence two solutions). Triange $OPA$ is a right triangle. Find $angleOPA$. Construct angle equal to $90-2 cdot OPA$ that has $P$ as vertex and $OP$ as one of the sides (again, there are two ways to construct the angle, choose wisely), The other side of the angle will intersect line $OA$ at some point, that point will be the center of $H$.
– Vasya
Jul 30 at 13:35
Hi, Thanks for your suggestion. The circle center of $H$ cannot be on the line $OA$ though. Your suggestion sets $r_H=3*r_D$ (if I'm not mistaken) but the radius $r_H$ is actually known and fixed.
– the_toast
Jul 30 at 14:11